The calculation of the other two terms is much more involved. Fortunately the expression for \(£_{m}^{G}\pi ^{ij}\)  is already well known [1]}[2]} and we merely quote the result,
\(£_{m}^{G}\pi ^{ij}_{x} &=-\frac{g^{1/2}}{2\kappa }\left(R^{ij}_{x}-\frac{1}{2}g^{ij}_{x}R_{x}\right)N_{xt}+ \frac{\kappa }{g^{1/2}_{x}} \, g^{ij}_{x}\left(\pi ^{kl}_{x}\pi _{klx}-\frac{1}{2}\pi ^{2}_{x}\right)N_{xt}\\&-\frac{4\kappa }{g^{1/2}_{x}}\left(\pi ^{ik}_{x}\pi ^{j}_{kx}-\frac{1}{2}\pi _{x}\pi ^{ij}_{x}\right)N_{xt}+\frac{g^{1/2}_{x}}{2\kappa }\left(\nabla ^{i}_{x}\nabla ^{j}_{x}N_{xt} - g^{ij}_{x}\nabla ^{k}_{x}\nabla _{kx}N_{xt}\right)~.\) 
